This is an extension of [1]}. It is proven that
\(f\in I(T^{h+1}B;h)\Leftrightarrow f^*\equiv B^* \,(\mathrm {mod}\,T^{n-h}).\) 
The result then follows because if \(f\)  runs over \(I(T^{h+1}B; h)\)  such that \(f(0)\ne  0\) , then \(f^*\)  runs over all polynomials of degree exactly \(n\)  satisfying \(f^*\equiv B^*\,(\mathrm {mod}\,T^{n-h})\) , and for these, \(\Lambda _k(f^*)=\Lambda _k(f)\) .
